We have $p ( x )= x ^{3}- ax ^{2}+6 x – a$

$\because $   Zero of $x – a$ is $a$.                                  $[\because x-a=0 \Rightarrow x=a]$

$\therefore $ $p ( a )=( a )^{3}- a ( a )^{2}+6( a )- a = a ^{3}- a ^{3}+6 a – a =0+5 a =5 a$

Thus, the required remainder $=5 a$